1.3 參考解答

6. (a) Let $z=x+iy$

   $|(x+1)+i(y-2)|=2$

   $\Rightarrow (x+1)^2+(y-2)^2=2^2$

   This is a circle of radius $2$, centered at $(-1+2i)=(-1,2)$.

   (b) Let $z=x+iy$, then $z+1=(x+1)+iy$

   $\text{Re}(z+1)=x+1$

   $x+1=0 \Rightarrow z=-1$

   It’s a vertical line pass through $(-1,0)$.

   (c) Let $z=x+iy$

   $|z+2i|=|x+i(y+2)| \leq 1$

   $\Rightarrow x^2+(y+2)^2 \leq 1$

   This is the closed disk of radius $1$, centered at $-2i=(0,-2)$.

   (d) Let $z=x+iy$, then $z-2i=x+i(y-2)$

   $\text{Im}(z-2i)=y-2<6$

   $\therefore y>8$

7. Let $z=x+iy$, $|z|=\sqrt{x^2+y^2}$, $|\text{Re}(z)|=|x|$, $|\text{Im}(z)|=|y|$.

   Show that $\sqrt{2}\sqrt{x^2+y^2} \geq |x|+|y|$.

   $(\sqrt{2}\sqrt{x^2+y^2})^2-( |x|+|y|)^2$

   $=2(x^2+y^2)-x^2-2|x||y|-y^2$

   $=x^2-2|x||y|+y^2$

   $=|x|^2-2|x||y|+|y|^2$

   $=(|x|-|y|)^2 \geq 0$

   i.e. $(\sqrt{2}\sqrt{x^2+y^2})^2 \geq ( |x|+|y|)^2$

   and $\because \sqrt{2}\sqrt{x^2+y^2} \geq 0$, $|x|+|y| \geq 0$

   Thus, $\sqrt{2}\sqrt{x^2+y^2} \geq |x|+|y|$.

   $\therefore \sqrt{2}|z| \geq |\text{Re}(z)|+|\text{Im}(z)|$.

17. Case 1: